I have got an question on computing conditional expection I was working on the following conditional expectation problem find $$\mathbb{E}[X-Y|2X+Y]$$

where $\begin{bmatrix}X \\ Y \end{bmatrix} \sim \mathcal{N}\bigg(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} \sigma_x^{2} & \rho\sigma_x\sigma_y\\ \rho\sigma_x\sigma_y & \sigma_y^{2} \end{bmatrix}\bigg)$

I have started with using the linearity of the conditional expectation $$\mathbb{E}[X-Y|2X+Y] = \mathbb{E}[X|2X+Y] - \mathbb{E}[Y|2X+Y]$$

Then for $\mathbb{E}[Y|2X+Y]$, let's define A = $\begin{bmatrix}0 & 1\\2 & 1\end{bmatrix}$

$\begin{align} \begin{bmatrix}Y \\ 2X +Y \end{bmatrix} = A* \begin{bmatrix}X \\ Y \end{bmatrix}&\sim \mathcal{N}\bigg(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, A\begin{bmatrix} \sigma_x^{2} & \rho\sigma_x\sigma_y\\ \rho\sigma_x\sigma_y & \sigma_y^{2} \end{bmatrix}A^{T}\bigg)\\&\sim \mathcal{N}\bigg(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} \sigma_y^{2} & \sigma_y^{2} +2\rho\sigma_x\sigma_y\\ \sigma_y^{2} +2\rho\sigma_x\sigma_y & 4\sigma_x^{2} +4\rho\sigma_x\sigma_y+\sigma_y^{2} \end{bmatrix}\bigg) \end{align}$

Finally we can use https://en.wikipedia.org/wiki/Multivariate_normal_distribution in order to get $\mathbb{E}[Y|2X+Y]$. This methodology just seems quite tedious and I am wondering if someone has a better way to tackle this problem !

Thanks a lot


1 Answer 1


First of all you can do this method for the pair $Z_1=(X-Y)$ and $Z_2=(2X+Y)$ directly, with $$A=\begin{bmatrix}1&-1\\2&1\end{bmatrix}$$

But, without doing this matrix multiplication, the only thing you need to compute is the following using the conditional mean formula will be (because $\mu_{Z_i}=0$): $$\mu_{Z_1|Z_2=z_2}={\operatorname{cov}(Z_1,Z_2)\over \operatorname{var}(Z_2)}z_2$$

which can be calculated as: $$\operatorname{var}(Z_2)=4\operatorname{var}(X)+\operatorname{var}(Y)+4\operatorname{cov}(X,Y)$$


  • $\begingroup$ thanks for you quick answer indeed the answer is almost direct in that case$$\sigma_{Z_1|Z_2=z_2}=\operatorname{var}(Z_1) - { \operatorname{cov}(Z_1,Z_2)^2\over \operatorname{var}(Z_2)}$$ $\endgroup$
    – ashu24
    Commented Nov 13, 2020 at 14:28

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